Why 'Enthalpy change' (Delta H) is equal to 'Heat transfer at constant pressure' (Qp)? [closed]

Why 'Enthalpy change' (∆H) is equal to 'Heat transfer at constant pressure' (Qp)?

∆H = ∆U + ∆pV, here only expansion work done by the sustem is added. If non expansion work is done on the system then it should be, ∆H ≠ Qp. What's wrong? Explain.

closed as unclear what you're asking by Mithoron, pentavalentcarbon, airhuff, Todd Minehardt, bonFeb 13 '18 at 13:27

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• Comments are not for extended discussion; this conversation has been moved to chat. – jonsca Feb 13 '18 at 23:03

In,Thermodynamics,Work is defined as
$W$= $-\int_{V_1}^{V_2}pdV$. So, if $dV$=0, there is no work done on the system,as $V_1=V_2$.
Coming to the enthalpy,you are right.
$\Delta$H=$\Delta$U + $\Delta$($PV$).But if you apply the product rule then,
$\Delta$H=$\Delta$U + $P$$\DeltaV + V$$\Delta$P.
At, constanst presurre,$\Delta$P=$0$. so at constant pressure,$\Delta$H=$\Delta$U + $P$$\DeltaV. But,according,to first law of Thermodynamics, Q= \DeltaU - W.At,constant pressure, the p in the integral comes out.So, W= -p\int_{V_1}^{V_2}dV= -p(V_2-V_1) = -p$$\Delta$V.So, first law in constant presuure becomes,
$Q_p$= $\Delta$U - (-$p$$\DeltaV)= \DeltaU+p$$\Delta$V= $\Delta$H. Thus, it is justified why Enthalpy change equals Heat change at constant pressure.

Enthalpy is a intensive property of the material that has nothing to do with the specific process that the material is subjected to. It just so happens in a constant pressure condition, involving only P-V work, that the change in enthalpy is equal to the heat added.

If non expansion work is done on the system, then ∆H≠Qp. It should be mentioned while writing '∆H = ∆U + p∆V =Qp', that 'it is applicable when only p-V work is done'.

Thanks to @ChesterMiller